If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field,then the orbit will be or The path executed by a charged particle whose motion is perpendicular to magnetic field is

This question has Statement $1$ and Statement $2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1$: $A$ charged particle is moving at a right angle to a static magnetic field. During the motion,the kinetic energy of the charge remains unchanged.
Statement $2$: $A$ static magnetic field exerts a force on a moving charge in the direction perpendicular to the magnetic field.

$A$ proton accelerated by a potential difference of $500 \ kV$ flies through a uniform transverse magnetic field of $0.1 \ T$. The field is spread over a region of $1.0 \ cm$ thickness. The angle through which the proton gets deviated from its original direction is (Proton mass $= 1.6 \times 10^{-27} \ kg$ and charge of proton $= 1.6 \times 10^{-19} \ C$) (in $rad$)

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

$A$ particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the field is

$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is